Question

The difference between the two acute angles of a right - angled triangle is $\frac{2\pi }{5}$ radians. Express the angles in degrees.

Answer 1

Let ${\theta }_{1}and{\theta }_2$ be two acute angles of a right angled triangle.

$\therefore$ different of acute angles

${\theta }_1-{\theta }_2=\frac{2\pi }{5}\text{radians}$

$\because$ in a right angled triangle,

${\theta }_1+{\theta }_2=\frac{\pi }{2}$

${\theta }_1-{\theta }_2=\frac{2\pi }{5}$ . . .(i)

${\theta }_1+{\theta }_2=\frac{\pi }{2}$ . . . (ii)

On solving

$2{\theta }_1=\frac{2\pi }{5}+\frac{\pi }{2}$

${\theta }_1=\frac{9\pi }{20}$

From equation (ii)

${\theta }_2=\frac{\pi }{20}$

So angles in degrees

${\theta }_1=\frac{9\pi }{20}\times \frac{180}{\pi }={81}^{\circ }$

and ${\theta }_2=\frac{\pi }{20}\times \frac{180}{\pi }=9^{\circ }$